1) A bubble of 6.00 mol helium is submerged at a certain depth in liquid water w

Question

1) A bubble of 6.00 mol helium is submerged at a certain depth in liquid water when the water ( and thus the helium ) undergoes a temperature increase T of 20°C at constant pressure. As a result, the bubble expands. (a) How much heat Q is added to the helium during the expansion and temperature increase? (b) What is the change Eint in the internal energy of the helium during the temperature increase? (c) How much work W is done by the helium as it expands against the pressure of the surrounding water during the temperature increase?

Explanation / Answer

(a)
Since the pressure is constant, the heat Q = nCPT,
where CP = CV +R for an ideal gas.
Therefore we have Q = n*(Cv + R)T
Q = 5* (5/2 * 8.31)*20
Q = 2493 J
Heat Q added during Expansion, Q = 2493 J

(b)
Change in Internal Energy , U = N*cV*T
Eint = 6 * 3/2 * 8.31 * 20
Eint = 1495.8 J

Change in Internal Energy, Eint = 1495.8 J

(c)
Relation bte Heat , Internal Energy & Work is given by
U = Q + W
W = U - Q
W = 1495.8  - 2493 J
W = -997.2 J

Work done by Helium , = - 997.2 J

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